The following lemma appears in Fields and Galois theory book by Patrick Morandi
If $K$ is a finite extension of $F$, then $K$ is algebraic and finitely generated over $F$
and in its proof uses the fact that if $\text{dim}_{F}(K)=n$ then for every $a\in{K}$ the set $\{1,a,\dots,a^n\}$ is linearity dependent.
And it isn't clear to me why that's true. I thought the reason was that the set $\{1,a,\dots,a^n\}$ have n+1, but for $a=1$ we have $\{1,a,\dots,a^n\}$=$\{1\}$. So I guess that the set always has not n elements. Any hint to see that?
The claim is NOT that $\{1, a, \dots, a^n\}$ consists of $n+1$ elements but that the set is linearly dependent (over the base field $F$). If $\{1, a, \dots, a^n\}$ has less than $n+1$ elements this is clear. Say we have $a^i=a^j$ for $i,j \le n$ and $i \ne j$. Then the non-trivial linear combination $$1 \cdot a^i + (-1) \cdot a^j =0$$ shows that the set is linearly dependent. If it has exactly $n+1$ elements - then the fact that dim$_F(K)=n$ gives that any set of $n+1$ or more elements is linearly dependent over $F$ meaning there is a non-trivial linear combination $$ c_0 + c_1a + c_2a^2 + \dots + c_na^n =0$$
In both cases we have a polynomial such that $f(a) = 0$ (first case: $f(x) = x^i-x^j$, second case $f(x) = c_0 + c_1x + \dots + c_nx^n$) - hence $a$ is algebraic over $F$ which was to be shown.